To construct a confidence interval for the mean expenditure on a child's last birthday gift, we can use the formula for the confidence interval for the mean when the sample size is relatively small (n < 30) or when the population standard deviation is not known. However, since the sample size is 25 and the data is roughly bell-shaped, we can use the t-distribution for this calculation.
Given:
- Sample size (n) = 25
- Mean (\(\bar{x}\)) = $31
- Standard deviation (s) = $5
- Confidence level = 99%
Step 1: Determine the t-value
For a 99% confidence level with n - 1 degrees of freedom (df = 25 - 1 = 24), we look up the t-value in the t-distribution table or use a calculator.
For df = 24 at a 99% confidence level, the t-value is approximately 2.797.
Step 2: Calculate the standard error (SE)
\[ SE = \frac{s}{\sqrt{n}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1 \]
Step 3: Calculate the margin of error (ME)
\[ ME = t \times SE = 2.797 \times 1 = 2.797 \]
Step 4: Construct the confidence interval
\[ \text{Lower limit} = \bar{x} - ME = 31 - 2.797 \approx 28.2 \] \[ \text{Upper limit} = \bar{x} + ME = 31 + 2.797 \approx 33.8 \]
Final Result
The 99% confidence interval for the mean amount spent on a child's last birthday gift is approximately:
\[ \boxed{(28.2, 33.8)} \]